What is the arc length of sqrt(2-x^2) – Calc II question?
To find the arc length ( L ) of the curve defined by ( y = \sqrt{2 - x^2} ) for the interval ( 0 \leq x \leq 1 ), please use the arc length formula. Additionally, you may verify your answer by recognizing that this curve represents a portion of a circle.
5 Answers
A simple way is to imagine that the equation is an equatin of a circle
y = √(2 – x²)
y² = 2 – x²
x² + y² = 2
with a center at the origin and radius of √2
when x = 0, y = √2
when x = 1, y = 1
The angle formed with center as the angle between the two points is 45 degrees, or in other words, the arc length is 45/360 of the circumference of the circle
arc length
= 45/360 x 2 π r
= (1/8)(2π)(√2)
≈ 1.1107 units
(a.) ??a million+(dx/dy)^2 dy=arc length dx/dy=a million/2(36-y^2)^(-a million/2)(-2y)=-y/?36… Arc length=??a million+y^2/(36-y^2) dy from 0 to 3 (b.) ignore approximately my integration calculations for now for the reason which you haven’t any longer found out this way of integration yet. only be sure my answer including your graphing utility. ?6/?36-y^2 dy 6arcsin(y/6) evaluated at limits: 6arcsin(a million/2)-6arcsin(0)=6pi/6=pi
Hi
The formula for arc-length is:
L = (a to b) ∫ [√(1 + (dy/dx)^2)] dx
a will be x = 0 and b will be x = 1. Since y = √(2 – x^2), we can find dy/dx to be:
y = √(2 – x^2)
dy/dx = (2 – x^2)’ * (1/2)(1/√(2 – x^2)) = -x/√(2 – x^2)
Now, we just plug that into the formula. You’ll have to use a trig-substitution to solve this one.
L = (0 to 1) ∫ [√(1 + (-x/√(2 – x^2))^2)] dx
= (0 to 1) ∫ [√(1 + ((x^2)/(2 – x^2))] dx
= (0 to 1) ∫ [√(((2 – x^2)/(2 – x^2)) + ((x^2)/(2 – x^2))] dx
= (0 to 1) ∫ [√((2 – x^2 + x^2)/(2 – x^2))] dx
= (0 to 1) ∫ [√(2/(2 – x^2))] dx
Let x = √(2)sin(t), then dx = √(2)cos(t) dt and the limits change to 0 to π/4
(0 to π/4) ∫ [√(2/(2 – (√(2)sin(t))^2))] √(2)cos(t) dt
= (0 to π/4) ∫ [√(2/(2 – 2sin(t)^2))] √(2)cos(t) dt
= (0 to π/4) ∫ [√(2/(2(1 – sin(t)^2)))] √(2)cos(t) dt
= (0 to π/4) ∫ [√(1/(1 – sin(t)^2))] √(2)cos(t) dt
= (0 to π/4) ∫ [√(1/cos(t)^2)] √(2)cos(t) dt
= (0 to π/4) ∫ [√(sec(t)^2)] √(2)cos(t) dt
= (0 to π/4) ∫ [sec(t)] √(2)cos(t) dt
= (0 to π/4) ∫ [√(2)sec(t)cos(t)] dt
= (0 to π/4) ∫ [√2] dt
= [t√2] (0 to π/4)
= (π/4)√(2) – (0)√(2)
= (π/4)√(2) – 0
= √(2)π/4
≈ 1.111
So the arc-length is about 1.111.
I hope this helps!
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